Axiom Of Choice

We have the following indirect implication of form equivalence classes:

109 \(\Rightarrow\) 131 given by the following sequence of implications, with a reference to its direct proof:

Implication	Reference
109 \(\Rightarrow\) 66	clear
66 \(\Rightarrow\) 67	Existence of a basis implies the axiom of choice, Blass, A. 1984a, Contemporary Mathematics
67 \(\Rightarrow\) 76	clear
76 \(\Rightarrow\) 131	clear

Here are the links and statements of the form equivalence classes referenced above:

Howard-Rubin Number	Statement
109:	Every field \(F\) and every vector space \(V\) over \(F\) has the property that each linearly independent set \(A\subseteq V\) can be extended to a basis. H.Rubin/J.~Rubin [1985], pp 119ff.
66:	Every vector space over a field has a basis.
67:	\(MC(\infty,\infty)\) \((MC)\), The Axiom of Multiple Choice: For every set \(M\) of non-empty sets there is a function \(f\) such that \((\forall x\in M)(\emptyset\neq f(x)\subseteq x\) and \(f(x)\) is finite).
76:	\(MC_\omega(\infty,\infty)\) (\(\omega\)-MC): For every family \(X\) of pairwise disjoint non-empty sets, there is a function \(f\) such that for each \(x\in X\), f(x) is a non-empty countable subset of \(x\).
131:	\(MC_\omega(\aleph_0,\infty)\): For every denumerable family \(X\) of pairwise disjoint non-empty sets, there is a function \(f\) such that for each \(x\in X\), f(x) is a non-empty countable subset of \(x\).

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